You have found the following ages (in years) of all 5 zebras at your local zoo: $ 22,\enspace 4,\enspace 40,\enspace 11,\enspace 8$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{22 + 4 + 40 + 11 + 8}{{5}} = {17\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $5$ years $25$ years $^2$ $4$ years $-13$ years $169$ years $^2$ $40$ years $23$ years $529$ years $^2$ $11$ years $-6$ years $36$ years $^2$ $8$ years $-9$ years $81$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{25} + {169} + {529} + {36} + {81}} {{5}} $ $ {\sigma^2} = \dfrac{{840}}{{5}} = {168\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{168\text{ years}^2}} = {13\text{ years}} $ The average zebra at the zoo is 17 years old. There is a standard deviation of 13 years.